Number of elements less than or equal to a given number in a given subarray

4.4

Given an array ‘a[]’ and number of queries q. Each query can be represented by l, r, x. Your task is to print the number of elements less than or equal to x in the subarray represented by l to r.

Examples:

Input : arr[] = {2, 3, 4, 5}
            q = 2
            0 3 5
            0 2 2 
Output : 4
         1
Number of elements less than or equal to
5 in arr[0..3] is 4 (all elements)

Number of elements less than or equal to
2 in arr[0..2] is 1 (only 2)

Naive approach The naive approach for each query traverse the subarray and count the number of elements which are in the given range.

Efficient Approach The idea is to use-Binary Index Tree.

Note in the following steps x is the number according to which you have to find the elements and the subarray is represented by l, r.
Step 1: Sort the array in ascending order.
Step 2: Sort the queries according to x in ascending order, initialize bit array as 0.
Step 3: Start from the first query and traverse the array till the value in the array is less than equal to x. For each such element update the BIT with value equal to 1
Step 4: Query the BIT array in the range l to r

// C++ program to answer queries to count number
// of elements smaller tban or equal to x.
#include<bits/stdc++.h>
using namespace std;

// structure to hold queries
struct Query
{
    int l, r, x, idx;
};

// structure to hold array
struct ArrayElement
{
    int val, idx;
};

// bool function to sort queries according to k
bool cmp1(Query q1, Query q2)
{
    return q1.x < q2.x;
}

// bool function to sort array according to its value
bool cmp2(ArrayElement x, ArrayElement y)
{
    return x.val < y.val;
}

// updating the bit array
void update(int bit[], int idx, int val, int n)
{
    for (; idx<=n; idx +=idx&-idx)
        bit[idx] += val;
}

// querying the bit array
int query(int bit[], int idx, int n)
{
    int sum = 0;
    for (; idx > 0; idx -= idx&-idx)
        sum += bit[idx];
    return sum;
}

void answerQueries(int n, Query queries[], int q,
                              ArrayElement arr[])
{
    // initialising bit array
    int bit[n+1];
    memset(bit, 0, sizeof(bit));

    // sorting the array
    sort(arr, arr+n, cmp2);

    // sorting queries
    sort(queries, queries+q, cmp1);

    // current index of array
    int curr = 0;

    // array to hold answer of each Query
    int ans[q];

    // looping through each Query
    for (int i=0; i<q; i++)
    {
        // traversing the array values till it
        // is less than equal to Query number
        while (arr[curr].val <= queries[i].x && curr<n)
        {
            // updating the bit array for the array index
            update(bit, arr[curr].idx+1, 1, n);
            curr++;
        }

        // Answer for each Query will be number of
        // values less than equal to x upto r minus
        // number of values less than equal to x
        // upto l-1
        ans[queries[i].idx] = query(bit, queries[i].r+1, n) -
                              query(bit, queries[i].l, n);
    }

    // printing answer for each Query
    for (int i=0 ; i<q; i++)
        cout << ans[i] << endl;
}

// driver function
int main()
{
    // size of array
    int n = 4;

    // initialising array value and index
    ArrayElement arr[n];
    arr[0].val = 2;
    arr[0].idx = 0;
    arr[1].val = 3;
    arr[1].idx = 1;
    arr[2].val = 4;
    arr[2].idx = 2;
    arr[3].val = 5;
    arr[3].idx = 3;

    // number of queries
    int q = 2;
    Query queries[q];
    queries[0].l = 0;
    queries[0].r = 2;
    queries[0].x = 2;
    queries[0].idx = 0;
    queries[1].l = 0;
    queries[1].r = 3;
    queries[1].x = 5;
    queries[1].idx = 1;

    answerQueries(n, queries, q, arr);

    return 0;
}

Output:

1
4

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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